In this article, we shall describe the construction of line segments and a few other constructions that involve the basic skill of constructing a line segment. Those are constructing the perpendicular bisector of a line segment, bisecting an angle, and constructing a triangle whose sides are in a given ratio with sides of another triangle. We have shown these constructions using a ruler and a pair of compasses only.

Learn All the Concepts on Line Segments

Construction of Line Segment Using Compass

In order to construct a line segment of a given length using a compass, we follow the given steps:

1. Obtain the length of the line segment.
2. Mark a point \(A\) (say) on the plane of the paper and draw a line, say \(l,\) passing through it.
3. Place the metal point of the compass at zero marks on the ruler and open out it such that the pencil point on the arc indicates the length of the line segment on the ruler.

4. Transfer the compass as it is to the line \(l\) so that the metal point is on \(A.\)
5. With the pencil, point makes a minor stroke on \(l\) to cut it as \(B.\)
6. The segment \(AB\) so obtained is the required segment of the given length.

Construct a Perpendicular Bisector of a Line Segment

Draw the perpendicular bisector of a given line segment \(AB\) of length \({\rm{5\,cm}}\) and the steps of construction are given below:

1. Draw the line segment \({\rm{AB = 5cm}}\)
2. With \(A\) as centre and radius, more than half \(AB\) draw arcs, one on each side of \(AB.\)
3. With \(B\) as a centre and the same radius as before, draw arcs, cutting the previously drawn arcs at \(P\) and \(Q.\)
4. Join \(PQ\) meeting \(AB\) at \(M.\) Then, \(PQ\) is the perpendicular bisector of \(AB.\)

Verification: Measure \(AM\) and \(MB.\) You would find that \(AM=MB\) 
Also, on measuring, we find that \(\angle PMB = \angle PMA = {90^ \circ }\)

Hence, \(PM\) is the perpendicular bisector of \(AB.\)

Draw a Line Perpendicular to a Given Line From a Point on It

A line \(XY\) is given, and \(P\) is a point on it. Draw a line through \(P\) perpendicular to \(XY.\) 

Step of Construction:

Let \(XY\) be the given line, and \(P\) be a point on it.

1. With centre \(P\) and any radius, draw a semicircle to intersect \(XY\) at \(A\) and \(B.\)
2. With centre \(A\) and any radius more than \(PA,\) draw an arc.
3. With centre \(B\) and the same radius, draw another arc, cutting the previously drawn arc at \(Q.\)
4. Join \(PQ.\)

Then, \(QP⊥XY\)

Verification: Measure \(∠QPX\) and \(∠QPY.\) You would find that \(\angle QPX = \angle QPY = {90^ \circ }\)

Draw a Line Perpendicular to a Given Line From a Point Outside It

A line \(XY\) is given, and \(P\) is a point outside it. Draw a line through \(P\) perpendicular to \(XY.\)

Steps of construction:

Let \(XY\) be the given line, and \(P\) be a point outside it.

1. With \(P\) as a centre and a convenient radius, draw an arc intersecting \(XY\) at \(A\) and \(B.\)
2. With \(A\) as a centre and a radius greater than \(\frac{1}{2}(A B)\) draw an arc.
3. With \(B\) as a centre and the same radius, draw another arc, cutting the previously drawn arc at \(Q.\)
4. Join \(PQ,\) meeting \(XY\) at \(L.\) Then, \(PL\) is the required perpendicular on \(XY.\)

Verification: Measure \(∠PLX\) and \(∠PLY\). You would find that \(\angle PLX = \angle PLY = {90^ \circ }\)

Draw a Line Parallel to a Given Line Through a Point Outside It

A line \(XY\) is given, and \(P\) is a point outside it. Draw a line through \(P\) parallel to \(XY.\)

Steps of construction:

Let \(XY\) be the given line, and \(P\) be a given point outside it.

1. Take any point \(Q\) on \(XY.\)
2. Join \(QP\)
3. Draw \(∠RPQ\) such that \(∠RPQ=∠PQY\) as shown in the figure.
4. Extend \(RP\) on both sides.

Then, the line \(RP\) passes through the point \(P\) and \(RP || XY.\)

Division of Line Segment Construction

In order to divide a line segment internally in a given ratio \(m:n\) where both \(m\) and \(n\) are positive integers, we follow the given steps:

1. Draw a line segment \(AB\) of a given length by using a ruler.
2. Draw any ray \(AX\) mark off \((m+n)\) points \(A_{1}, A_{2}, \ldots, A_{m}, A_{m+1}, \ldots A_{m+n}\) such that \(A A_{1}=A_{1} A_{2}=A_{m+n-1} A_{m+n}\)
3. Join \(B A_{m+n}\)
4. Through the point \(A_{m}\) draw a line parallel to \(A_{m+n} B\) by making an angle equal to \(\angle A A_{m+n} B\) at \(A_{m}\) Suppose this line meets \(AB\) at a point \(P.\)

The point \(P\) so obtained is the required point which divides \(AB\) internally in the ratio \(m:n.\)

Justification: In \(\Delta A B A_{m+n}\), we observe that \(A_{m} P\) is parallel to \(A_{m+n} B\) Therefore, by the basic proportionality theorem, we have

\(\frac{A A_{m}}{A_{m} A_{m+n}}=\frac{A P}{P B}\)
\(\Rightarrow \quad \frac{A P}{P B}=\frac{m}{n}\)
[\(\because \,\,\,\frac{{A {A_m}}}{{{A_m}{A_{m + n}}}} = \frac{m}{n}\) (By construction)]
\( \Rightarrow AP:PB = m:n\)

Hence, \(P\) divides \(AB\) in the ratio \(m:n.\)

Dividing a Given Line Segment in a Given Ratio

Divide a line segment of length \({\rm{10cm}}\) internally in the ratio \(3:2.\) We follow the steps for the construction:

1. Draw a line segment \(A B=10 \mathrm{~cm}\) by using a ruler.
2. Draw any ray making an acute angle \(\angle B A X\) with \(AB.\)
3. Along \(AX,\) mark off \(5(=3+2)\) points \(A_{1}, A_{2}, A_{3}, A_{4}\) and \(A_{5}\) such that \(A A_{1}=A_{1} A_{2}=A_{2} A_{3}=A_{3} A_{4}=A_{4} A_{5}\)
4. Join \(B A_{5}\)
5. Through \(A_{3}\) draw a line \(A_{3} P\) parallel to \(A_{5} B\) by making an angle equal to \(\angle A A_{5} B\) at \(A_{3}\) intersecting \(AB\) at a point \(P.\) The point \(P\) so obtained is the required point, which divides \(AB\) internally in the ratio \(3:2.\)

Solved Examples – Construction of Line Segment

Q.1. Draw a line segment of length 6.6 cm. Bisect it and measure the length of each part.
Ans: We follow the given steps to construct the line segment:
1. Draw a line segment \(A B=6.6 \mathrm{~cm}\) by using a graduated ruler.
2. With \(A\) as centre and radius more than half of \(AB,\) draw arcs, one on each side of \(AB.\)
3. With \(B\) as the centre and the same radius as in step \(2,\) draw arcs cutting the arcs drawn in step \(2\) at \(E\) and \(F,\) respectively.
4. Draw the line segment with \(E\) and \(F\) as endpoints. Suppose it meets \(AB\) at \(M.\) Then, \(M\) bisects the line segment \(AB.\)
By measuring \(AM\) and \(MB,\) we find that \(A M=M B=3.3 \mathrm{~cm}\)

Q.2. Draw a line segment PQ of length 5.6 cm. Draw the perpendicular bisector of this line segment.
Ans: We follow the following steps for constructing the perpendicular bisects of \(PQ:\)

1. Draw a line segment \(PQ=5.6 \mathrm{~cm}\) by using a graduated ruler.
2. With \(P\) as centre and radius more than half of \(PQ,\) draw two arcs, one on each side of \(PQ.\)
3. With \(Q\) as a centre and the same radius as in step II, draw arcs cutting the arcs drawn in the previous step at \(L\) and \(M,\) respectively.
4. Draw the line segment with \(L\) and \(M\) as endpoints. The line segment \(LM\) is the required perpendicular bisector of \(PQ.\)

Q 3. Construct a triangle similar to a given triangle ABC with its sides equal to \(\frac{3}{4}\) of the corresponding sides of the triangle ABC (i.e., of scale factor \(\frac{3}{4}\)
Ans: Given a triangle \(ABC,\) we are required to construct another triangle whose sides are \(\frac{3}{4}\) of the corresponding side of the triangle \(ABC.\)

Steps of construction:

1. Draw any ray \(BX\) making an acute angle with \(BC\) on the side opposite to the vertex \(A.\)
2. Locate \(4\) (the greater of \(3\) and \(4\) in \(\frac{3}{4}\) points \(B_{1}, B_{2}, B_{3}\) and \(B_{4}\) on \(BX\) so that \(B B_{1}=B_{1} B_{2}=B_{2} B_{3}=B_{3} B_{4}\)
3. Join \(B_{4} C\) and draw a line through \(B_{3}\) (the \({{\rm{3}}^{{\rm{rd}}}}\) point, \(3\) being smaller of \(3\) and \(4\) in \(\frac{3}{4}\) parallel to \(B_{4} C\) to intersect \(BC\) at \(C’.\)
4. Draw a line through \(C’\) parallel to the line \(CA\) to intersect \(BA\) at \(A’\). Then \(∆ A’BC’\) is the required triangle.
\(\frac{B C^{\prime}}{C^{\prime} C}=\frac{3}{1}\)
Therefore, \(\frac{B C}{B C^{\prime}}=\frac{B C^{\prime}+C C}{B C^{\prime}}=1+\frac{C^{\prime} C}{B C^{\prime}}=1+\frac{1}{3}=\frac{4}{3}\), i.e, \(\frac{B C^{\prime}}{B C}=\frac{3}{4}\)
Also,  \(C’A\) is parallel to \(CA.\) Therefore, \(\Delta A’BC’ \sim \Delta ABC\) (Why?)
So, \(\frac{A^{\prime} B}{A B}=\frac{A^{\prime} C^{\prime}}{A C}=\frac{B C^{\prime}}{B C}=\frac{3}{4}\)

Q.4. Construct a perpendicular bisector of a line segment of length 12 cm using a compass.
Ans: Steps of construction:
1. Draw a line segment \(AB\) of length \(12 \mathrm{~cm}\)
2. With centre \(A\) and radius more than half of \(AB,\) draw arcs, one on each side of \(AB.\)
3. With \(B\) as a centre and the same radius as before, draw arcs, cutting the previously drawn arcs at \(E\) and \(F\) respectively.
4. Join \(EF\) intersecting \(AB\) at \(M.\) Then \(M\) bisects the line segment \(AB\) as shown in the given diagram.
Verification: Measure \(AM\) and \(MB.\) You will find that \(AM=MB.\)

Q.5. Draw a line segment PQ of length 8.4 cm. Draw the perpendicular bisector of this line segment.
Ans: We follow the following steps for the construction of the perpendicular bisector of \(PQ\)
1. Draw a line segment \(P Q=8.4 \mathrm{~cm}\) by using a graduated ruler.
2. With \(P\) as centre and radius more than half of \(PQ,\) draw two arcs, one on each side of \(PQ.\)
3. With \(Q\) as a centre and the same radius as in step II, draw arcs cutting the arcs drawn in the previous step at \(L\) and \(M\) respectively.
4. Draw the line segment with \(L\) and \(M\) as endpoints. The line segment \(LM\) is the required perpendicular bisector of \(PQ.\)

Summary

In this article, first, we discussed the construction of line segments using the compass. Then, we discussed the construction of a perpendicular bisector of a line segment which covered how to draw a line perpendicular to a given line from a point on it, outside it and how to draw a line parallel to a given line through a point outside it along with examples.

After that, we discussed the division of line segment. We have also provided the solved examples along with a few FAQs.

Learn All the Concepts on Practical Geometry

FAQs

Q.1. How to construct a line segment of a given length using a compass?
Ans: To construct a line segment of a given length using a compass, we follow the given steps:
1. Obtain the length of the line segment.
2. Mark a point \(A\) (say) on the plane of the paper and draw a line, say \(l,\) passing through it.
3. Place the metal point of the compasses at zero marks on the ruler and open out it such that the pencil point on the ark indicates the length of the line segment on the ruler.
4. Transfer the compasses as it is to the line \(l\) so that the metal point is on \(A.\)
5. With the pencil, point makes a minor stroke on \(l\) to cut it as \(B.\)
6. The segment \(AB\) so obtained is the required segment of the given length.

Q.2. What is a line segment that has measure and direction?
Ans: A line segment has two endpoints. You can measure the length of a segment but not of a line. A ray is a part of a line with one endpoint and goes on infinitely in only one direction. You cannot measure the length of a ray.

Q.3. Which is the measure of a line segment?
Ans: Unlike the line, a line segment has a definite length. The length of the line segment can be measured in units like metric units such as millimetres, centimetres, or customary units like feet or inches.

Q.4. Which is the best way of measuring the line segment?
Ans: When measuring the length of the line segment, place the endpoint of the segment on the zero marks on the ruler, then see where it ends. 

Q.5. What is the line segment of the circle?
Ans: The line segment joining any two points on the circumference of a circle is called its chord. The line segment that crosses the circle by passing through its centre is the diameter. The diameter is double the length of the radius, and it is the longest chord of a circle.